Results 1 to 9 of 9

Math Help - simplify this function

  1. #1
    Newbie woohoo's Avatar
    Joined
    Dec 2008
    Posts
    20

    Question simplify this function

    <br /> <br />
\ln y \;=\;11\bigg[\frac{x^2}{2} - \frac{11}{2}\ln(x^2+11)\bigg] + C<br />

    can someone rewrite this in terms of y?

    I know you need to multiply by e, but I don't know what to do from there.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Dec 2008
    Posts
    111
    Dear woohoo,

    not multiplying.

    Tipp: if ln(y) = C than y = e^C
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie woohoo's Avatar
    Joined
    Dec 2008
    Posts
    20
    K so I put it to the power of e then.

    But I'm not quite sure how to simplify it after that (ie. with the + and the - and how to simplify into x and division)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    It's not "multiplying", but composing with the exponential function.
    \exp(x)=e^x

    \frac{1}{11} ~ \ln y=\frac{x^2}{2}-\frac{11}{2} ~ \ln(x^2+11)+\frac{C}{11}

    \ln \left(y^{1/11}\right)=\frac{x^2}{2}-\ln \left(\left(x^2+11\right)^{11/2}\right)+\frac{C}{11}

    y^{1/11}=\exp \left(\frac{x^2}{2}-\ln \left(\left(x^2+11\right)^{11/2}\right)+\frac{C}{11}\right)

    y^{1/11}=\frac{\exp \left(\frac{x^2}{2}+\frac{C}{11}\right)}{\exp\left  (\ln \left(\left(x^2+11\right)^{11/2}\right)\right)}

    y^{1/11}=\frac{\exp \left(\frac{x^2}{2}\right) \cdot \exp \left(\frac{C}{11}\right)}{(x^2+11)^{11/2}}


    And honestly, I don't think it is possible to write this in terms of y
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2008
    Posts
    111
    Note:

    e^{a+b} = e^a*e^b

    But I don't believe you will get much simplier form.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie woohoo's Avatar
    Joined
    Dec 2008
    Posts
    20
    Quote Originally Posted by Moo View Post
    And honestly, I don't think it is possible to write this in terms of y

    That's what I'm starting to think, as well. =/
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie woohoo's Avatar
    Joined
    Dec 2008
    Posts
    20
    *Forgot to mention this, but C is an arbitrary constant.

    ie 3C=C

    It's value doesn't matter.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Dec 2008
    From
    Indiana
    Posts
    127
    It does look like there not too much you can do to simplify this, except the piece with ln.

    \ln y \;=\;11\bigg[\frac{x^2}{2} - \frac{11}{2}\ln(x^2+11)\bigg] + C<br />


    y = e^{\frac{11}{2}x^2+C}*e^{\frac{-121}{2}\ln (x^2+11)}
    y = e^{\frac{11}{2}x^2+C}*e^{\ln (x^2+11)^\frac{-121}{2}}
    y =\bigg[(x^2+11)^\frac{-121}{2}\bigg]*e^{\frac{11}{2}x^2+C}
















    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie woohoo's Avatar
    Joined
    Dec 2008
    Posts
    20
    Yeah, that didn't work.

    But it makes sense to me.

    Any more takers? = /
    Last edited by woohoo; December 7th 2008 at 10:20 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simplify Boolean Function
    Posted in the Calculus Forum
    Replies: 9
    Last Post: June 9th 2010, 12:59 PM
  2. Simplify this function
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: April 17th 2009, 02:22 PM
  3. Can I simplify the following function
    Posted in the Calculus Forum
    Replies: 9
    Last Post: April 5th 2009, 07:58 PM
  4. Simplify Inverse Trig Function
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: December 2nd 2008, 03:43 AM
  5. simplify this function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 28th 2007, 11:13 AM

Search Tags


/mathhelpforum @mathhelpforum